four numbers

the powerpoint is here

thanks to Mike Askew for this question

it explores oddness and evenness
and divisibility by 4, including - possibly - a divisibility test for 4

and it can be set in the context of probability
finding all the outcomes, theoretically
maybe checking these by experiment

1 to 9 multiplied

the idea for this (and the first question) is from nrich 1134

thinking enables the (usually unique) solutions to be found

also see: NRICH multiplication equation sudoku

cancelling down

divisibility rules

divisibility tests DIY

using ~ to mean 'is a factor of'.
this work involves simple algebra and some knowledge of multiplication tables

divisibility test for 7

 what's happening here?







suppose you want a divisibility test for 7
split the number into the number of Tens (T) and the number of Units (U).
7 ~ 10T + U
7 ~ 7T + 3T + U [technique: splitting into the divisor + some extra ]
7 ~ 3T + U
so multiply the number of tens by 3 and add the number of units
if this divides by 7 so does the whole number
check: 133 : 3 x 13 + 3 =42 which is in the 7 x table so, so is 133
(42 could be continued: 3 x 4 + 2 = 14)

this isn't a particularly helpful technique for 7
looking for a better one:
it is good to have near decades in the 7 times table
there are two or three: 49, 21, 91 to choose from
which works best?

using 49:
7 ~ 10T + U
7 ~ 10T + U + 49U
7 ~ 10(T + 5U)
7 ~ T + 5U
so if 7 is a factor of T + 5U then it is a factor of the original number
check: 91 : 9 + 5 x 1 = 14.

using 21:
7 ~ 10T + U
7 ~ 10T + U - 21U
7 ~ 10(T - 2U)
7 ~ T - 2U
So if 7 is a factor of T - 2U then it is a factor of the original number
check: 357 : 35 - 2 x 7 = 21.

This (latest) test is claimed to have originated in Russia
and can be extended to larger numbers, e.g. 6951:
695 - 2 = 693
69 - 6 = 63
6 - 6 = 0, is a M(7) so 6951 is a multiple of 7.

using 91:
7 ~ 10T + U
7 ~ 10T + U - 91U
7 ~ 10(T - 9U)
7 ~ T - 9U
7 ~ T - 7U - 2U
7 ~ T - 2U (as before)

David Wilson offers this planar graph to test for divisibility by 7:
write down a number n
start at the small white node at the bottom of the graph
for each digit d in n, follow d black arrows in a succession, and as you move from one digit to the next, follow 1 white arrow

for example, if n = 325, follow 3 black arrows, then 1 white arrow, then 2 black arrows, then 1 white arrow, and finally 5 black arrows

if you end up back at the white node, n is divisible by 7.



divisibility test for 13

39 could be helpful:
13 ~ 10T + U
13 ~ 10T + U + 39U
13 ~ 10(T + 4U)
so if T + 4U is a multiple of 13 so is the original number

or, using 91:
13 ~ 10T + U
13 ~ 10T + U - 91U
13 ~ 10(T - 9U)
13 ~ T - 9U
so if T - 9U is a multiple of 13 so is the original number

establish these tests (or better still, find your own)

• T + 2U for 19
• T + 3U for 29
• T - 3U for 31
• T - 5U for 17
• 2T + U for 8
• 2T - U for 12
• T + 7U for 23
• T + 10U for 33

and of course,

• T + U for 9 
• T - U for 11

some related questions:
What is the largest three-digit multiple of 7?
What is the largest three-digit multiple of 13?
What is the largest three-digit multiple of 19?